# Question Video: Powers in terms of Multiple Angles from Eulerβs Formula Mathematics

1. Use Eulerβs formula to express sinΒ³ π cosΒ² π in the form π sin π + π sin 3π + π sin 5π, where π, π, and π are constants to be found. 2. Hence, find the solutions of sin 5π β sin 3π = 0 in the interval 0 β©½ π < π.

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### Video Transcript

Use Eulerβs formula to express sin cubed π cos squared π in the form π sin π plus π sin three π plus π sin five π, where π, π, and π are constants to be found. Hence, find the solutions of sin five π minus sin three π equals zero in the interval π is greater than or equal to zero and less than π. Give your answer in exact form.

We begin by recalling the fact that sin π is equal to one over two π times π to the ππ minus π to the negative ππ. And cos π is equal to a half times π to the ππ plus π to the negative ππ. This means we can find the product of sin cubed π and cos squared π. We can write it as one over two π times π to the ππ minus π to the negative ππ cubed times a half times π to the ππ plus π to the negative ππ squared. One over two π cubed is negative one over eight π. And a half squared is one-quarter. So we can rewrite our expression a little bit further. We find the product of negative one over eight π and a quarter. And we get negative one over 32π. And we can rewrite the rest of our expression as shown.

Weβre now going to use the binomial theorem to expand each of the sets of parentheses. The first part becomes π to three ππ plus three choose one π to the two ππ times negative π to the negative ππ and so on. And this simplifies to π to the three ππ minus three π to the ππ plus three π to the negative ππ minus π to the negative three ππ. Letβs repeat this process for π to the ππ plus π to the negative ππ squared. When we do, we get π to the two ππ plus two π to the zero which is just two plus π to the negative two ππ.

Weβre going to need to find the product of these two expressions. Weβll need to do that really carefully. Weβll need to ensure that each term in the first expression is multiplied by each term in the second expression. And we can write sin cubed π cos squared π as shown. Now, thereβs quite a lot going on here. So you might wish to pause the video and double-check your answer against mine. Weβre going to gather the corresponding powers of π together.

Weβll gather π to the five ππ and π to the negative five ππ. Weβll collect π to the plus and minus three ππ. And weβll gather π the ππ and π to the negative ππ. Letβs neaten things up somewhat. We end up with negative one over 32π times π to the five ππ minus π to the negative five ππ minus π to the three ππ minus π to the negative three ππ minus two times π to the ππ minus π to the negative ππ. And now, you might be able to spot why we chose to do this. We can now go back to the given formulae. Letβs clear some space for the next step.

We kind of unfactorize a little. And we can rewrite sin cubed π cos squared π as shown. And we can therefore replace π to the ππ plus π to the negative ππ with sin π and so on. And we can see that sin cubed π cos squared π is equal to a 16th times two sin π plus sin three π minus sin five π. Since π, π, and π are constants to be found, we can say that π, the coefficient of sin π, is one-eighth. π, the coefficient of sin three π, is a 16th. And π, the coefficient sin five π, is negative one 16th.

Letβs now consider part two of this question. We begin by using our answer to part one and multiplying both sides by 16. We then subtract two sin π from both sides and multiply through by negative one. And we can now see that weβve got an equation in sin five π minus sin three π. Weβre told that sin five π minus sin three π is equal to zero. So we let two sin π minus 16 sin cubed π cos squared π be equal to zero. And then, we factor by two sin π. Since the product of these two terms is equal to zero, this means that either of these terms must be equal to zero. So either two sin π is equal to zero and dividing by two, we could see that sin π is equal to zero or one minus eight sin squared π cos squared π is equal to zero. Given the interval π is greater than or equal to zero and less than π, we can see that one of our solutions is when π is equal to zero.

Weβre going to rewrite our other equations somewhat. We know that sin two π is equal to two sin π cos π. Squaring this, we get sin squared two π equals four sin squared π cos squared π. And thatβs in turn means that our equation is one minus two sin squared two π equals zero. Rearranging to make sin two π the subject, we see that sin two π is equal to plus or minus one over root two. Starting with the positive square root for π in the interval given, we know that sin two π is equal to one over root two when π is equal to π by eight or three π by eight. Similarly, we can solve for the negative square root. And we get five π by eight and seven π by eight. And there are therefore five solutions to the equation sin five π minus sin three π equals zero in the interval π is greater than or equal to zero and less than π. They are zero π by eight, three π by eight, five π by eight, and seven π by eight.